Once the switch S is closed, although the initial current through the inductor is 0, the change in current through it is maximal. Recalling that , one realizes that the voltage across must be non-zero at the start. Thus, plots (C), (D), and (E) are eliminated.

Now, one must decide between choices (A) and (B). Once the circuit reaches equilibrium, i.e., the elements reach their asymptotic values. Specifically, the voltage at A goes to 0, since the inductor has no potential difference across it (to wit ). Once the switch is opened, the current suddenly changes, and , thus the inductor has a voltage across it, and the voltage at A becomes nonzero. Because of the diode, this would necessarily have to be a negative voltage. Since there is only to dissipate the voltage, the magnitude of the voltage once the switch is opened should be bigger than that initially, when the switch is closed at . Choose (B).